Find $\lim_{x\to 1}\dfrac{x\ln(x)}{x^2-1}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac14$ (Choice B) B $4$ (Choice C) C $\dfrac12$ (Choice D) D The limit doesn't exist.
Explanation: Substituting $x=1$ into $\dfrac{x\ln(x)}{x^2-1}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 1}\dfrac{x\ln(x)}{x^2-1} \\\\ &=\lim_{x\to 1}\dfrac{\dfrac{d}{dx}[x\ln(x)]}{\dfrac{d}{dx}[x^2-1]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 1}\dfrac{\ln(x)+x\cdot\dfrac1x}{2x} \\\\ &=\dfrac{\ln(1)+1}{2(1)} \gray{\text{Substitution}} \\\\ &=\dfrac12 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 1}\dfrac{\dfrac{d}{dx}[x\ln(x)]}{\dfrac{d}{dx}[x^2-1]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{x\ln(x)}{x^2-1}=\dfrac12$.